The given question states that you have to proof that $\mathbb{R} [X]/(x^2+1)$ and $\mathbb{R} [x]/(x^2+3)$ are isomorphic, and then give an explicit isomorphism between them.
I have already showed that are isomorphic, given that both rings are isomorphic to $\mathbb{C}$.
However i don't understand the second part (that is, giving the explicit isomorphism), which was given by our prof.
We construct two morphisms:
$\phi$ : $\mathbb{R} [X]$ $\rightarrow$ $\mathbb{R} [X]$, and $\pi$ : $\mathbb{R} [X]$ $\rightarrow$ $\mathbb{R} [X] / (x^3 + 3)$
given by $x \rightarrow \frac{x}{\sqrt{3}}$ and $\frac{x}{\sqrt{3}} \rightarrow \frac{\overline{x}}{\sqrt{3}}$. Now, it can easily be seen that $\phi$ is an isomorphism, and that $\pi$ is exhaustive.
Now, i can't see why he states that ker($ \pi \circ \phi $)= ($x^2+1$).
When i "plug", for example, $x^2+1$, i evaluate the composition doing:
$x^2+1 \rightarrow \frac{x^2+1}{\sqrt{3}} \rightarrow \frac{\overline{x^2+1}}{\sqrt{3}} \rightarrow \frac{\overline{-3+1}}{\sqrt{3}} \rightarrow \frac{\overline{-2}}{\sqrt{3}} $, which is not $\overline{0}$.
I searched from similar questions here, but i still can't understand this specific example. Did i misunderstood something in the way? Any help would be greatly apreciated!
The map $\phi$ doesn't send any polynomial $f(x)$ to $\frac{f(x)}{\sqrt{3}}$. It sends specifically the element $x$ to $\frac{x}{\sqrt{3}}$, and the images of the other elements are determined by this. We have:
$\phi(x^2+1)=\phi(x)^2+\phi(1)=(\frac{x}{\sqrt{3}})^2+1=\frac{x^2}{3}+1,$
and this is indeed a multiple of $x^2+3$. This shows that $(x^2+1)\subseteq\ker(\pi\circ\phi)$. Since $(x^2+1)$ is a maximal ideal of $\mathbb{R}[x]$, it follows that the two ideals are equal.